How to plot an angle (theta) vs Current for the following question

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I am new to Matlab.
I wanted to find the Torque given as T= A*I*cos(theta) - B*I.^2*sin(2*thetha)
with A= 0.01 B= 0.001; I varies from 0 to 100 and thetha from 0 to 45.
I want to find the maximum T (Torque) and plot I vs Theta
This my basic syntax:
A=0.01; %Wb is the unit of A
B=0.001; %H is the unit of Inductance
I=(0:10:100);
theta=(0:1:45);
T=A*I.Cos(theta)-B*I.^2*Sin(2*theta);

采纳的回答

Walter Roberson
Walter Roberson 2021-7-19
A=0.01; %Wb is the unit of A
B=0.001; %H is the unit of Inductance
I =(0:10:100); %row!
theta=(0:1:45).'; %column!
T = A .* I .* cosd(theta) - B .* I.^2 .* sind(2*theta);
surf(I, theta, T)
xlabel('I'); ylabel('theta'); zlabel('Torque')
[maxT, idx] = max(T(:));
[r,c] = ind2sub(size(T), idx);
best_theta = theta(r);
best_I = I(c);
fprintf('best torque is %g at theta = %g and I = %g\n', maxT, best_theta, best_I);
best torque is 1 at theta = 0 and I = 100
  7 个评论
Walter Roberson
Walter Roberson 2021-7-20
Yes, using meshgrid. You can avoid using ind2sub() -- but you still end up using linear indexing.
A=0.01; %Wb is the unit of A
B=0.001; %H is the unit of Inductance
I_vec =(0:10:100);
theta_vec = (0:1:45);
[I, theta] = meshgrid(I_vec, theta_vec);
T = A .* I .* cosd(theta) - B .* I.^2 .* sind(2*theta);
surf(I, theta, T)
xlabel('I'); ylabel('theta'); zlabel('Torque')
[maxT, idx] = max(T(:));
best_theta = theta(idx);
best_I = I(idx);
fprintf('best torque is %g at theta = %g and I = %g\n', maxT, best_theta, best_I);
best torque is 1 at theta = 0 and I = 100

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